Answer: 0.4 moles
Explanation:
Given that:
Volume of gas V = 11L
(since 1 liter = 1dm3
11L = 11dm3)
Temperature T = 25°C
Convert Celsius to Kelvin
(25°C + 273 = 298K)
Pressure P = 0.868 atm
Number of moles N = ?
Note that Molar gas constant R is a constant with a value of 0.00821 atm dm3 K-1 mol-1
Then, apply ideal gas equation
pV = nRT
0.868atm x 11dm3 = n x (0.00821 atm dm3 K-1 mol-1 x 298K)
9.548 atm dm3 = n x 24.47atm dm3mol-1
n = (9.548 atm dm3 / 24.47atm dm3 mol-1)
n = 0.4 moles
Thus, there are 0.4 moles of the gas.
Answer:
a) 88.48%
b) 0.05625 mol
Explanation:
2CH₃CH₂OH(l) → CH₃CH₂OCH₂CH₃(l) + H₂O(g) Reaction 1
CH₃CH₂OH(l) → CH₂═CH₂(g) + H₂O(g) Reaction 2
a) CH₃CH₂OH = 46.0684 g/mol
CH₃CH₂OCH₂CH₃ = 74.12 g/mol
1 mol CH₃CH₂OH ______ 46.0684 g
x ______ 50.0 g
x = 1.085 mol CH₃CH₂OH
1 mol CH₃CH₂OCH₂CH₃ ______ 74.12 g g
y ______ 35.9 g
y = 0.48 mol CH₃CH₂OCH₂CH₃
100% yield _____ 0.5425 mol CH₃CH₂OCH₂CH₃
w _____ 0.48 mol CH₃CH₂OCH₂CH₃
w = 88.48%
b) Only 0.96 mol of ethanol reacted to form diethyl ether. This means that 0.125 mol of ethanol did not react. 45% of 0.125 mol reacted to form ethylene. Therefore, 0.05625 mol of ethanol reacted by the side reaction (reaction 2). Since 1 mol of ethanol leads to 1 mol of ethylene, 0.05625 mol of ethanol produces 0.05625 mol of ethylene.
<span>CH</span>₃<span>CH</span>₂<span>COOH + H</span>₂<span>O </span>↔ <span> CH</span>₃<span>CH</span>₂<span>COO</span>⁻<span> + H</span>₃<span>O</span>⁺<span>
</span>
pH = 0.5 pKa + 0.5 pCa
0.5 pCa = pH - 0.5 pKa
= 4.2 - (0.5 * (-log 1.34 x 10⁻⁵)) = 1.76
pCa = 3.53
Ca = antilog - 3.52 = 3 x 10⁻⁴
where Ca is the acid concentration
H2SO3 or sulfurous acid is actually a strong acid. We know
for a fact that strong acids completely dissociate into its component ions in a
solution, that is:
<span>H2SO3 --> 2H+ + SO3-</span>
<span>So from the equation above, there are 2 moles of H+</span>
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